Find the optimal solution using dual simplex algorithm
$begingroup$
Maximize $ z=-5x_1+10x_2+8x_3$
subject to constraints
$3x_1+5x_2+2x_3<=60$
$4x_1+4x_2+4x_3>=72$
$x_1<=0$
$x_1,x_2,x_3>=0$
Thanks for help :)
optimization linear-programming simplex
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|
show 9 more comments
$begingroup$
Maximize $ z=-5x_1+10x_2+8x_3$
subject to constraints
$3x_1+5x_2+2x_3<=60$
$4x_1+4x_2+4x_3>=72$
$x_1<=0$
$x_1,x_2,x_3>=0$
Thanks for help :)
optimization linear-programming simplex
$endgroup$
$begingroup$
At which step did a problem occur? One important hint: For $=$ and $geq$ constraints you need an artificial variable each.
$endgroup$
– callculus
Jan 30 at 21:21
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In the initial table on row all values are not$ <=0$ but i do not know what to do next :(
$endgroup$
– Corn of Doom
Jan 30 at 21:30
$begingroup$
It´s not clear for me what do you mean by your comment. Please rephrase it. Something what is obvious is that $x_1=0$ if you combine the constraints $x_1leq 0$ and $x_1geq 0$. Therefore you can drop the terms with $x_1$ at the objetive function and the constraints.
$endgroup$
– callculus
Jan 30 at 21:35
$begingroup$
The best you can do is to show your initial table.
$endgroup$
– callculus
Jan 30 at 21:39
$begingroup$
One second pls :)
$endgroup$
– Corn of Doom
Jan 30 at 21:43
|
show 9 more comments
$begingroup$
Maximize $ z=-5x_1+10x_2+8x_3$
subject to constraints
$3x_1+5x_2+2x_3<=60$
$4x_1+4x_2+4x_3>=72$
$x_1<=0$
$x_1,x_2,x_3>=0$
Thanks for help :)
optimization linear-programming simplex
$endgroup$
Maximize $ z=-5x_1+10x_2+8x_3$
subject to constraints
$3x_1+5x_2+2x_3<=60$
$4x_1+4x_2+4x_3>=72$
$x_1<=0$
$x_1,x_2,x_3>=0$
Thanks for help :)
optimization linear-programming simplex
optimization linear-programming simplex
asked Jan 30 at 21:12
Corn of DoomCorn of Doom
315
315
$begingroup$
At which step did a problem occur? One important hint: For $=$ and $geq$ constraints you need an artificial variable each.
$endgroup$
– callculus
Jan 30 at 21:21
$begingroup$
In the initial table on row all values are not$ <=0$ but i do not know what to do next :(
$endgroup$
– Corn of Doom
Jan 30 at 21:30
$begingroup$
It´s not clear for me what do you mean by your comment. Please rephrase it. Something what is obvious is that $x_1=0$ if you combine the constraints $x_1leq 0$ and $x_1geq 0$. Therefore you can drop the terms with $x_1$ at the objetive function and the constraints.
$endgroup$
– callculus
Jan 30 at 21:35
$begingroup$
The best you can do is to show your initial table.
$endgroup$
– callculus
Jan 30 at 21:39
$begingroup$
One second pls :)
$endgroup$
– Corn of Doom
Jan 30 at 21:43
|
show 9 more comments
$begingroup$
At which step did a problem occur? One important hint: For $=$ and $geq$ constraints you need an artificial variable each.
$endgroup$
– callculus
Jan 30 at 21:21
$begingroup$
In the initial table on row all values are not$ <=0$ but i do not know what to do next :(
$endgroup$
– Corn of Doom
Jan 30 at 21:30
$begingroup$
It´s not clear for me what do you mean by your comment. Please rephrase it. Something what is obvious is that $x_1=0$ if you combine the constraints $x_1leq 0$ and $x_1geq 0$. Therefore you can drop the terms with $x_1$ at the objetive function and the constraints.
$endgroup$
– callculus
Jan 30 at 21:35
$begingroup$
The best you can do is to show your initial table.
$endgroup$
– callculus
Jan 30 at 21:39
$begingroup$
One second pls :)
$endgroup$
– Corn of Doom
Jan 30 at 21:43
$begingroup$
At which step did a problem occur? One important hint: For $=$ and $geq$ constraints you need an artificial variable each.
$endgroup$
– callculus
Jan 30 at 21:21
$begingroup$
At which step did a problem occur? One important hint: For $=$ and $geq$ constraints you need an artificial variable each.
$endgroup$
– callculus
Jan 30 at 21:21
$begingroup$
In the initial table on row all values are not$ <=0$ but i do not know what to do next :(
$endgroup$
– Corn of Doom
Jan 30 at 21:30
$begingroup$
In the initial table on row all values are not$ <=0$ but i do not know what to do next :(
$endgroup$
– Corn of Doom
Jan 30 at 21:30
$begingroup$
It´s not clear for me what do you mean by your comment. Please rephrase it. Something what is obvious is that $x_1=0$ if you combine the constraints $x_1leq 0$ and $x_1geq 0$. Therefore you can drop the terms with $x_1$ at the objetive function and the constraints.
$endgroup$
– callculus
Jan 30 at 21:35
$begingroup$
It´s not clear for me what do you mean by your comment. Please rephrase it. Something what is obvious is that $x_1=0$ if you combine the constraints $x_1leq 0$ and $x_1geq 0$. Therefore you can drop the terms with $x_1$ at the objetive function and the constraints.
$endgroup$
– callculus
Jan 30 at 21:35
$begingroup$
The best you can do is to show your initial table.
$endgroup$
– callculus
Jan 30 at 21:39
$begingroup$
The best you can do is to show your initial table.
$endgroup$
– callculus
Jan 30 at 21:39
$begingroup$
One second pls :)
$endgroup$
– Corn of Doom
Jan 30 at 21:43
$begingroup$
One second pls :)
$endgroup$
– Corn of Doom
Jan 30 at 21:43
|
show 9 more comments
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$begingroup$
At which step did a problem occur? One important hint: For $=$ and $geq$ constraints you need an artificial variable each.
$endgroup$
– callculus
Jan 30 at 21:21
$begingroup$
In the initial table on row all values are not$ <=0$ but i do not know what to do next :(
$endgroup$
– Corn of Doom
Jan 30 at 21:30
$begingroup$
It´s not clear for me what do you mean by your comment. Please rephrase it. Something what is obvious is that $x_1=0$ if you combine the constraints $x_1leq 0$ and $x_1geq 0$. Therefore you can drop the terms with $x_1$ at the objetive function and the constraints.
$endgroup$
– callculus
Jan 30 at 21:35
$begingroup$
The best you can do is to show your initial table.
$endgroup$
– callculus
Jan 30 at 21:39
$begingroup$
One second pls :)
$endgroup$
– Corn of Doom
Jan 30 at 21:43